I am confused with the statement that "consider the augmentation map $\epsilon_L :\mathfrak{U}(L) \rightarrow \mathbb{F}$ which is the unique algebra homomorphism induced by $\epsilon_{L}(x)=0 , \forall x \in L$ (I hope $L$ here is seen as the copy $T^1$ in $\mathfrak{U}(L)$) where $\mathfrak{U}(L)$ is the universal enveloping algebra of the Lie algebra $L$ and the kernel of $\epsilon_{L}$ is said to be the augmentation ideal. My confusion is that if $\epsilon_{L}(x)=0 , \forall x \in L$ then does not that imply $ker(\epsilon_{L})$ is just $\mathfrak{U}(L)/\mathbb{F}$?
Augmentation Ideal of Universal Enveloping Algebra
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abstract-algebra
representation-theory
lie-algebras
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0$\ker \epsilon_L=\bigoplus_{n\geq 1}T^n(L)$. – 2017-02-15
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0It should NOT be $T^n(L)$ assuming $T^i(L)s $ are the tensor powers of $L$.Rather it might be $\ker \epsilon_L=\bigoplus_{n\geq 1} \mathfrak{U}_{n}(L)$ where $\{ \mathfrak{U}_{i}(L) \}$ is the obvious grading. Actually I did not like the description of the ideal in terms of $kernel$ of the map. One can just say The ideal $ \bigoplus_{n\geq 1} \mathfrak{U}_{n}(L)$ is the augmentation ideal. It is an ideal is obvious. – 2017-02-16
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0You are right. That was a silly error (apparently I have a hang-over from answering a question on the relationship between the tensor algebra and the enveloping algebra). The point is that the kernel is **not** the quotient. – 2017-02-16
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1The augmentation map makes sense without knowing that the Lie algebra _injects_ to the enveloping algebra, is one little point. And then its kernel is undeniably an ideal. That kernel is a sub-object of the enveloping algebra, not a quotient. Similarly, it is not clear a-priori (before Poincare-Birkhoff-Witt) that the sum of the images of $\bigotimes^n \mathfrak g$ is direct, etc. But that info is not needed to describe the augmentation ideal. – 2017-02-24
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You wanted to write this $\mathfrak U(L)\backslash\mathbb F$, right? Since it does not make sense to write $\mathfrak U(L)/\mathbb F$, because $\mathbb F$ not is an ideal of $\mathfrak U(L)$. The correct is $\ker \epsilon_L=\mathfrak U(L)\backslash\mathbb F^*$, this is direct by definition of $\epsilon_L$. Moreover, you can show that $\mathfrak U(L)=\mathbb F\oplus\ker\epsilon_L$ if you consider $\mathbb F$ is the subalgbera of $\mathfrak U(L)$ generated by $1.$